Problem: Simplify the following expression: $\dfrac{45r}{99r^2}$ You can assume $r \neq 0$.
Explanation: $ \dfrac{45r}{99r^2} = \dfrac{45}{99} \cdot \dfrac{r}{r^2} $ To simplify $\frac{45}{99}$ , find the greatest common factor (GCD) of $45$ and $99$ $45 = 3 \cdot 3 \cdot 5$ $99 = 3 \cdot 3 \cdot 11$ $ \mbox{GCD}(45, 99) = 3 \cdot 3 = 9 $ $ \dfrac{45}{99} \cdot \dfrac{r}{r^2} = \dfrac{9 \cdot 5}{9 \cdot 11} \cdot \dfrac{r}{r^2} $ $\phantom{ \dfrac{45}{99} \cdot \dfrac{1}{2}} = \dfrac{5}{11} \cdot \dfrac{r}{r^2} $ $ \dfrac{r}{r^2} = \dfrac{r}{r \cdot r} = \dfrac{1}{r} $ $ \dfrac{5}{11} \cdot \dfrac{1}{r} = \dfrac{5}{11r} $